The equations of the circle which pass throug | vedclass

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Question

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle passes through the origin $(0, 0)$,we have $c = 0$. The circle

Vedclass · Accepted Answer

$x^2+y^2 \pm 4x \pm 8y=0$

Vedclass · Answer

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle passes through the origin $(0, 0)$,we have $c = 0$. The circle makes an intercept of length $4$ on the $x$-axis,so $2\sqrt{g^2 - c} = 4$. Since $c = 0$,$2|g| = 4$,which implies $g = \pm 2$. The circle makes an intercept of length $8$ on the $y$-axis,so $2\sqrt{f^2 - c} = 8$. Since $c = 0$,$2|f| = 8$,which implies $f = \pm 4$. Substituting $g = \pm 2$ and $f = \pm 4$ into the equation $x^2 + y^2 + 2gx + 2fy = 0$,we get $x^2 + y^2 \pm 4x \pm 8y = 0$.

The equations of the circle which pass through the origin and make intercepts of lengths $4$ and $8$ on the $x$ and $y$-axes respectively are

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